Q.Given: \[ \theta + \phi = \frac{7\pi}{12},\quad \tan\theta = \sqrt{3} \Rightarrow \theta = \frac{\pi}{3} \] \[ \phi = \frac{7\pi}{12} - \frac{\pi}{3} = \frac{7\pi}{12} - \frac{4\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \] \[ \tan\phi = \tan\left(\frac{\pi}{4}\right) = 1 \] (a) \(\cfrac{1}{2}\) (b) 1 (c) \(\cfrac{1}{\sqrt3}\) (d) \(\cfrac{\sqrt3}{2}\)
Answer: B
\[ \tan\theta = \sqrt{3} = \tan\left(\frac{\pi}{3}\right) \Rightarrow \theta = \frac{\pi}{3} \] \[ \text{Since } \theta + \phi = \frac{7\pi}{12} \Rightarrow \frac{\pi}{3} + \phi = \frac{7\pi}{12} \Rightarrow \phi = \frac{7\pi}{12} - \frac{\pi}{3} \Rightarrow \phi = \frac{7\pi - 4\pi}{12} = \frac{3\pi}{12} = \frac{3 \times 180^\circ}{12} = 45^\circ \] \[ \therefore \tan\phi = \tan 45^\circ = 1 \]
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